I'm not sure what they're getting at with the 4j either. I'd say ignore it.

I have to run to class, so I'll try to answer as quick as possible.

I did the following to figure out the Thevenin equivalent:

1) Remove the load (Zl)

2) Make the input voltage a short.

3) Now you are looking for Rth seen from the right side of the circuit. In redrawing the circuit, you can see that 5ohms and j(ohms) are in parallel. So far we have (5 || j). Considering this to be one impedance (say, Zparallel), and redrawing the circuit, you can see that Rth is actually Zparallel in series with the 3(j) inductor.

Therefore, Rth = 3j + (5 || j)

4) Now Vth is the voltage seen at the load terminal with the load removed. So we have a voltage divider between the 5ohm resistor and 1(j) inductor, as the voltage seen at the far end of the 3(j) inductor is the same as the node between 5 and (1)j since there's no load Zl.

The voltage at the voltage divider node is then V(1j/(5+1j))

So your Thevenin equivalent is V(1j/(5+1j)) in series with 3j + (5 || 1j) in series with the original load impedance, Zl.